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IR  Graphical Interpretation Sample Question [#permalink]
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09 Mar 2013, 15:09
Hi, the OA for the first part is 4,0005,000. can anyone explain why? thanks
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Re: IR  Graphical Interpretation Sample Question [#permalink]
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10 Mar 2013, 04:41
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Hi ,
I am not sure if this is the most elegant way to do this..
Since we want approx answers we can assume that the part of the graph from point A to C is a straight line.
Point A = ( 12,900) Point C = ( 20,5100)
Slope = (5100900) / ( 2012) = 525
Hence Equation of line = (y900) = 525 ( x12)
Simplifying we get y= 525x  5400
Now we have to get sum of weights from ages 13 till 19 . Substituting for x we get
525( 13+ 14+ ...19 )  7*5400 525 (112) 37800 = 21000
From 21 to 30 we can assume average mass as 5300 factoring for a very slight increase from age 20 30 . Hence total weight is 5300* 10
Thus Total weight = 900+ 21000+ 5100+53000= 80,000
Hence average = 80,000/19 = 4210
Regards
Ganesh



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Re: IR  Graphical Interpretation Sample Question [#permalink]
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10 Mar 2013, 07:12
ganeshiyer1983 wrote: Now we have to get sum of weights from ages 13 till 19 . Substituting for x we get
525( 13+ 14+ ...19 )  7*5400 525 (112) 37800 = 21000
Thanks Ganesh! Could you please explain why you multiply 5400 by 7? I understand this is a number of all values for x from 13 to 19, but I don't get why you add it to the formula. thanks



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Re: IR  Graphical Interpretation Sample Question [#permalink]
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10 Mar 2013, 07:25
Hi,
The equation is y=525x 5400
for x=13 it becomes y= 525 ( 13) 5400 for x= 14 it becomes y= 525 ( 14) 5400 for x= 15 it will be y= 525 ( 15) 5400 Similarly continuing till x=19 and we will have 525 ( 19)  5400
Since we have to find the sum ,when we add all these up there will a total of 7 expressions and each of them will have the term 5400
So we could take 525 as a common factor and we will have 5400 seven times which can be expressed as 525 ( 13+14+15+...19)  (5700 *7)
Does this make sense?



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10 Mar 2013, 07:28
Yes, it does! thanks!



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Re: IR  Graphical Interpretation Sample Question [#permalink]
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10 Mar 2013, 16:02
Hi, if you give the graph a closer look, you will see that : Between 12 and 21, there are 10 values And between 21 and 30, there are 10 values [The set of ages is an evenly spaced set i.e the average equals the median, which is equal to 21] So, you can pick 1 value from the former and 1 value from the latter and average the corresponding two values of mass. For point B (16, 3000) and another point let's call it D ( 25, 5500) , the average mass is : (3000 + 5500)/2= 4250, which is between 4000 and 5000.. This holds true for any values selected from the previous intervals because the graph is incremental Hope that helps !
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10 Mar 2013, 20:03
Rock750 wrote: Hi,
if you give the graph a closer look, you will see that :
Between 12 and 21, there are 10 values
And between 21 and 30, there are 10 values [The set of ages is an evenly spaced set i.e the average equals the median, which is equal to 21]
So, you can pick 1 value from the former and 1 value from the latter and average the corresponding two values of mass.
For point B (16, 3000) and another point let's call it D ( 25, 5500) , the average mass is : (3000 + 5500)/2= 4250, which is between 4000 and 5000..
This holds true for any values selected from the previous intervals because the graph is incremental
Hope that helps ! This is what I did in the first place, but I picked A (12, 900) and D (25, 5500) and came up with the average mass of 3200 (900+5500)/2. why didn't it work? thanks



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Re: IR  Graphical Interpretation Sample Question [#permalink]
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11 Mar 2013, 01:03
LinaNY wrote: This is what I did in the first place, but I picked A (12, 900) and D (25, 5500) and came up with the average mass of 3200 (900+5500)/2. why didn't it work? thanks Hi LinaNY, You should take B from the former , it's more representative than any other value picked from the first intervall because it seems to be the middle point i.e the average of the first intervall while for the latter, it seems to have a constant corresponding mass of almost all the values picked from 21 to 30 > 5500 . I know it's tricky but when it's about IR, always try to pick the easiest values i.e the most logical ones. Hope that helps !
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Re: IR  Graphical Interpretation Sample Question [#permalink]
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06 May 2013, 07:49
Just by looking at the graph you can see that weights > 5000 account for 1/3 of the age range. With the Median weight at age 16 being 3000 lbs, you can see that the mean will be much higher than 3000 lbs. See this excellent article for how to determine this without calculations. comparemeanandmedianinlessthan20seconds146191.html
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Re: IR  Graphical Interpretation Sample Question [#permalink]
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12 Jan 2018, 09:33
Question 1 is test of weighted average concept.
for age 1219, average weight is roughly 3000 lbs for age 2030, average weight is roughly 5500 lbs
3000<11><8>5500 8 (1220)11(2030)
weighted average: (8/19) * 2500 + 3000 = (1200)approx + 3000 = 4200 lbs




Re: IR  Graphical Interpretation Sample Question
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